Find the set of circles with a given radius that pass through a given point

Question

Given a point $(x_0,y_0)$ and a radius $r$, how do you find the set of all circles that have that radius that pass through the point?

Let $(h,k)$ represent the center of the set of circles. Then, it's clear that we have the following, because we want the set of points $(h,k)$ that have a distance $r$ from $(x_0,y_0)$ such that the circle centered at $(h,k)$ has a radius of $r$:

$$ (x-h)^2 + (y-k)^2 = r^2$$

$$ (h-x_0)^2 + (k-y_0)^2 = r^2$$

It's unclear how to proceed from here. While it's true that we have two variables in two equations, and we must solve for $h$ and $k$, the algebra is really messy and I'm not entirely sure this problem admits a closed form solution. Any insights?

Answer 1

We are interested in circles in the $\langle x,y\rangle$ plane, therefore they will have Cartesian equation:

$$ (x-x_C)^2 + (y-y_C)^2 = r^2 $$

where their radius $r>0$ is fixed, so it remains to understand how to determine the centers.

In particular, given that we want the circles in question all to pass through $(x_P,\,y_P)$ and at the same time we have radius $r > 0$, it's evident that $(x_C,\,y_C)$ must belong to the circle with center $(x_P,\,y_P)$ and radius $r > 0$, that is:

$$ \begin{cases} x_C = x_P + r\,\cos(u) \\ y_C = y_P + r\,\sin(u) \\ \end{cases} \quad \quad \text{with} \; u \in [0,\,2\pi)\,. $$

In this way, we have determined the Cartesian equation of the sheaf of circles obtainable according to the chosen value of $u \in [0,\,2\pi)$; below a simple simulation in Mathematica:

{xP, yP, r, du} = {1, 2, 3, 2 Pi / 10};
{xC, yC} = {xP, yP} + r {Cos[u], Sin[u]};
circles = Table[(x - xC)^2 + (y - yC)^2 == r^2, {u, 0, 2 Pi - du, du}];
max = Max[Abs[{xP - 2 r, xP + 2 r, yP - 2 r, yP + 2 r}]];
ContourPlot[Evaluate[circles], {x, -max, max}, {y, -max, max},
            Axes -> True, AxesLabel -> {"x", "y"},
            Frame -> False, GridLines -> Automatic]

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